Question

Andy spent Rs. 450 in fencing the parallelogram shaped land (leaving 1 m for the gate) that he had before. If Andy would start from a point 5 m away from A (on side AB) and take the shortest possible path to the opposite side, he would reach exactly at D. As you know Land Grabbers restructured Andy’s land to be a rectangle having the same base and same width. Find the cost of fencing the new land in the same way at Rs. 6 /m.

Answer 1

Let us redraw the figure given like this. With what is given in the question, we can say that DE is the height of the parallelogram.

Let the unknown side of the parallelogram be ‘$x$’ m.

Perimeter of the parallelogram field which is fenced = $\frac{Totalcostoffencing}{Costoffencingfor1m}$ Rs. $\frac{450}{6}$ = $75$ m ( 1 mark)

So,

Perimeter = $(2\times 25+2x-1)=75$

=$2x–1=25$

$x=13$ m (1 mark)

So the other side of parallelogram is 13 m.

In $△ADE$, $AD$ = 13 m, $AE$ = 5 m

Using Pythagoras theorem,

$A{D}^2=A{E}^2+D{E}^2$

or, $D{E}^2={13}^2-5^2=144={12}^2$

So, $DE=12$ m (1 mark)

The restructured land is in the shape of a rectangle with same base and width.

So, length of the rectangular land = 25 m

Breadth of the rectangular land = 12 m

Perimeter = 2 (l + b) = 2 (25 + 12) = 2 $\times$ 37 m = 74 m (0.5 mark)

Perimeter to be fenced = (74 - 1) m = 73 m $(\because 1miskeptopenforentrytothefarm)$

Cost of fencing = Rs. $\frac{6}{m}$ × 73 m = Rs. 438 (0.5 mark)